emails and mathematical topics

Will I or will I not get flu?

Suddenly I got emails about mathematical topics from two of my friends. I have induced AP to program baroque numbers, which he was doing on and off, now on again. He uses the so-called genetic approach which is like simulated annealing + one additional kind of moves, where from two vertices one gets a new one (it simulates sexual reproduction). He has a better software, better computer, everything better than me but the results. Somehow it's not easy to communicate on long distance, in an irregular way, and it's hard for me to pass to him my experience. Actually, I am rusty these days myself, and will have to start almost from scratch (almost) if at all.

As the luck has it, around the same time I get an email from JBrz, who is excited about abc again. He and JBro have published a nice paper anbout abc years ago, about the integer and polynomial versions. So, they are serious, while I had only an amatourish interest in the topic, and only in the classical, integer version. But JBrz is all about the polynomial version these days. I am a social being, so I will try to join him.

I have a thousand of topics for knols (instead of working on one topic only), and now I have a thousand and two. And so it goes.

Recently I have finished ("published") my first knol about logarithmic and exponential functions, log and exp--a constructive and an axiomatic approaches. It has scored its first six views. The logarithmic knol is partially based on two knols about Integration of monotone functions and The ground level properties of integral.

Baroque conjectures

Let me address the views (guesses) on the four questions, about the perfect and baroque numbers, listed in the previous post:

• everybody believes that there are infinitely many (even) perfect numbers, because everybody believes that there are infinitely many Mersenne prime numbers p (i.e. prime numbers p such that 2^p-1 is a prime too);

• it seems to me that some (many? most?) specialists believe that there are only finitely many baroque numbers which are not perfect; some even think that all baroque non-perfect numbers are already known (no way!). Myself, I believe that there are infinitely many of them but finitely many for each baroqueness coefficient;

• nobody believes that there is any odd perfect number;

• I think that nobody believes that there is any odd baroque number; I don't either but I am not so sure :-).

I am interested in the race toward finding more baroque numbers.

Back to baroque numbers?

The classical, standard name for my term baroque numbers is pluperfect or polyperfect
numbers or similar. A natural number n (n = 1 2 ...) is called baroque when the sum sd(n) of all natural divisors of n is divisible by n; let's call

brq(n) := sd(n)/n

the baroqueness of n. When a baroqueness of a number is two, brq(n) = 2, then it is called perfect. The smallest perfect number, known since the ancient times, is n=6. Indeed, in this case

sd(n) = 1+2+3+4+6 = 12 = 2*n

Questions about the baroque numbers are among the oldest, open (unresolved) problems of the whole mathematics. They are difficult and, today, somewhat isolated from the main developments, hence they are not as popular among the best mathematicians as they used to be in the past, when Euclid, Fermat, Decart, Euler and others were interested in them. Let me list the main open questions:

• are there infinitely many perfect numbers?
• are there infinitely many baroque numbers?
• does there exist an odd perfect number?
• does there exist an odd baroque number > 1?

Of course n=1 is the only number for which brq(n) = 1.

The Euclid-Euler tandem (:-) proved that an even natural number n is perfect if and only if there exists a (unique) natural number p such that the following two conditions hold:

• 2^p - 1 is prime;
• n = 2^p-1*(2^p-1)

It is well known and easy to see that when 2^p-1 is prime then so must be the exponent p itself.

Another old result (an ancient observation) is that brq(120) = 3, i.e. 120 is a baroque number, and its baroqueness is 3.

a step beyond Euclid and Fermat, 6

Let  f : Z --> Z  be a polynomial function, i.e. f belongs to Z[X]. Then  f  is a residue obeying funtion, or  rof  for short, in the following sense:

• if  x=y mod n   then   f(x) = f(y) mod n

for every integer x y and natural n.

Now let  f  be an arbitrary residue obeying function. Let's define sequence x(0) x(1) ... satisfy the following iterative equality:

• x(n+1) = f(x(n))

for every   n=0 1...

Then x(0) x(1)... is a prime residue obeying sequence, or pros for short, in the following sense:

• if   x(i) = x(j) mod p   then   x(i+1) = x(j+1) mod m

for every i j = 0 1... and prime p=2 3 5 7 11...

Thus every pros mod p is eventually periodic:

indeed, let sequence x(0) x(1)... be a pros, and let  p  be a prime; then there are nonegative integers  i < j ≤ p  such that  x(i) = x(j) mod p.  Then starting with the index  i,  sequence x(n) is periodic mod p:

• x(n+d) = x(d) mod p

for   d := j-i,   and for every   n=i i+1...

This means that if prime  p  didn't divide any of the terms  x(0) ... x(j-1)  then  p  does not divide any term of the infinite sequence  x(0) x(1)....

a step beyond Euclid and Fermat, 5

For the sake of reference, you may like to open the previous note, step 4, in a separate window.

Let

Par(n)  :=  { k : 0 ≤ k ≤ n and k = n mod 2}

be the set of all non-negative numbers k ≤ n, which have the same parity as n; e.g.

• Par(0) = {0}
• Par(1) = {1}
• Par(2} = {0 2}
• Par(3) = {1 3}
• Par(4) = {0 2 4}

etc.

It follows from the initial values of the sequences (g(n) and (A(n), and from the last equality of "step 4" that:

• A(n)  =  ∏_{k ∈ Par(n)} g(k)     ∀ n=1 2...
  

The first three terms, g(0) g(1) g(2), of sequence (g(n)), are pairwise relatively prime.

If every two of the terms g(0) ... g(n) (with different indices) are relatively prime then obviously A(n-1) and A(n) are relatively prime too. Since, by the very definition, g(n+1) is the difference of A(n) and A(n-1), g(n+1) is relatively prime with A(n-1) and with A(n), which means that g(n+1) is relatively prime to every previous term g(0) ... g(n). This holds for every n=2 3 ... Thus every two different terms of sequence (g(n)) are relatively prime.

a step beyond Euclid and Fermat, 4

I've introduced the Euclid sequence e(n), and Fermat introduced his Fermat sequence f(n). Now let me introduce sequence g(n), which again has pairwise relatively prime terms. In another post I will show that sequence g is essentially different from sequences e and f and from all sequences euf. It is also satisfying that the terms g(n) are asymptotically smaller than e(n) and f(n), that they are of the order of the square root of e(n) or of f(n).

First, let me introduce an auxiliary sequence A(n):

• A(1) := 3
• A(2) := 10
• A(n+1) := A(n-1) ⋅ (A(n) - A(n-1))     ∀ n= 2 3 ...

so that A(3) = 3*(10-3) = 21, A(4) = 10*(21-10) = 110, etc.

It is not too hard to see that

A(n) = n mod 2     ∀ n=1 2 ...

Now let

• g(0) := 2
• g(1) := 3
• g(2) := 5
• g(n+1) := A(n) - A(n-1)     ∀ n=2 3 ...

so that g(3) = 7, g(4) = 11, g(5) = 89, etc. Also:

• A(n+1) := A(n-1) ⋅ g(n+1)     ∀ n= 2 3 ...

It is a fascinating topic to me but I am tired now. I'll continue later.

a step beyond Euclid and Fermat, 3

Let's make our first step beyond Euclid and Fermat. Their sequences are so similar that they beg to be placed under a roof of a common generalization. Let a b be two integers. Let's define:

• euf_{a b}(0)  :=  a+b;
• euf_{a b}(n+1)  :=  (euf_{a b}(n) - b) ⋅ euf_{a b}(n) + b
  

for every n=0 1...

Then

• euf_{a b}(n+1)  =  b + ∏_k=0...n euf_{a b}(k)
  

for every n=0 1...

Of course, if  B := euf_{a b}(k) - a  for certain non-negative integer k then:

• euf_{a B}(n)  =  euf_{a b}(k+n)

for every n=0 1...

For special values of integers a b, the sequence (euf_{a b}(n) : n=0 1...) becomes the Euclid sequence, when (a b) := (1 1), or the Fermat sequence, when (a b) := (2 3); when b=1 then we suppress  b  by writing

euf_a(n)  :=  euf_{a 1}(n)
Thus:

• e(n) = euf₁(n) — the Euclid sequence;
• f(n) = euf₂(n) — the Fermat sequence.

On the other hand, when a:=0, we obtain the constant sequence of values b.

The following properties of an Euclid-Fermat sequence euf_{a b} are equivalent:

• integers a b are relatively prime, i.e. gcd(a b) = 1;
• euf_{a b}(k) and euf_{a b}(n) are relatively prime whenever k and n are different.

We may rewrite the above simple recursive formula, which expresses euf_{a b}(n+1) in terms of euf_{a b}(n) as follows:

• euf_{a b}(n+1)  =  (euf_{a b}(n) - b/2)² + b - b²/4

This formula allows to study the rate of increase (or decrease) of an Euclid-Fermat sequence.

a step beyond Euclid and Fermat, 2

Fermat has defined his numbers:

• f(n) := 2²ⁿ + 1 for every n=0 1 ...

hoping that all of them are prime. Following Euler, we will see in another entry to this blog that this is not so. At this time let's just compare the Euclid numbers e(n) with Fermat numbers f(n). Surprisingly, they are quite similar in more than one way:

• f(0) = 3
• f(n+1) = (f(n) - 2) ⋅ f(n) + 2
• f(n+1) = 2 + ∏_k=0...n f(k)
  

for every n=0 1...

Polya has used the last formula to partially vindicate Fermat's hope—as in the case of the Euclidean sequence, also every two different Fermat numbers are relatively prime (hence once again we see that there are infinitely many different prime numbers). It follows that:

• p(n) ≤ f(n-1)     for every n=1 2...
  

---

The iterative formula for  f(n+1)  can be rewritten equivalently as follows:

• f(n+1) = (f(n) - 1)² + 1

for every n=0 1...

a step beyond Euclid and Fermat, 1

The eternally elegant Euclid's proof of infinitude of the set of prime numbers can be rephrased as follows: let

• e(0)  :=  2
  
• e(n+1)  :=  (e(n) - 1) ⋅ e(n) + 1

Then  (e(n) : n=1 2...)  is an increasing sequence of natural numbers such that each of its two different terms are relatively prime.

Indeed,

• e(n+1)  :=  1 + ∏_k=1...n e(k)

for every  n = 0 1 ...

Let  (p(k) : k=0 1 ...)  be the increasing sequence of all prime numbers. We see that

• p(n)  ≤  e(n)   for every n=0 1...

In fact:

• p(0) = e(0) = 2
• p(1) = e(1) = 3

but

• p(2) = 5 < 7 = e(2)

and

• p(n)  <  e(n)   for every n=2 3...